package com.ky.leetcode.剑指Offer;

import java.util.Stack;

/*

定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 O(1)。

    示例:
    MinStack minStack = new MinStack();
    minStack.push(-2);
    minStack.push(0);
    minStack.push(-3);
    minStack.min();   --> 返回 -3.
    minStack.pop();
    minStack.top();      --> 返回 0.
    minStack.min();   --> 返回 -2.

    提示：
    各函数的调用总次数不超过 20000 次

*/
public class 剑指Offer30包含min函数的栈 {

    public static void main(String[] args) {
        MinStack minStack = new MinStack();
        minStack.push(-2);
        minStack.push(0);
        minStack.push(-3);
        System.out.println(minStack.min());   //--> 返回 -3.
        minStack.pop();
        System.out.println(minStack.top());   //--> 返回 0.
        System.out.println(minStack.min());   //--> 返回 -2.
    }

    static class MinStack {

        private Stack<Integer> stack;
        private Stack<Integer> minStack;

        /**
         * initialize your data structure here.
         */
        public MinStack() {
            stack = new Stack<>();
            minStack = new Stack<>();
        }

        public void push(int x) {
            if (stack.isEmpty())
                minStack.push(x);
            stack.push(x);
            if (minStack.peek() >= x) {
                minStack.push(x);
            }
        }

        public void pop() {
            if (stack.isEmpty()) return;
            int pop = stack.pop();
            if (minStack.peek() == pop) {
                minStack.pop();
            }
        }

        public int top() {
            return stack.peek();
        }

        public int min() {
            return minStack.peek();
        }
    }
}
